Convert 5/2 to a mixed number first:
Divide 5 by 2; 2 will go into five 2 times with a remainder of 1. This means the whole number portion of the mixed number is 2, and the remainder of the fraction is 1; this gives us 2 1/2.
To plot 2 1/2, plot a point halfway between 2 and 3.
No, it's not possible for the sides of a triangle to have those lengths.
According to the triangle inequality theorem, the sum of any two sides of the triangle has to be bigger than the last side. Let's test this.

This inequality satisfies the triangle inequality theorem.

This also satisfies the theorem.

Uh oh. This does not satisfy the triangle inequality theorem. Thus, it is not possible for a triangle to have these side lengths.
Answer:
See the argument below
Step-by-step explanation:
I will give the argument in symbolic form, using rules of inference.
First, let's conclude c.
(1)⇒a by simplification of conjunction
a⇒¬(¬a) by double negation
¬(¬a)∧(2)⇒¬(¬c) by Modus tollens
¬(¬c)⇒c by double negation
Now, the premise (5) is equivalent to ¬d∧¬h which is one of De Morgan's laws. From simplification, we conclude ¬h. We also concluded c before, then by adjunction, we conclude c∧¬h.
An alternative approach to De Morgan's law is the following:
By contradiction proof, assume h is true.
h⇒d∨h by addition
(5)∧(d∨h)⇒¬(d∨h)∧(d∨h), a contradiction. Hence we conclude ¬h.
Answer:
e equels MC squared
Step-by-step explanation: